Explain conduction in solids in terms of the movement of free (delocalised) electrons in metallic conductors.
1. The Free‑Electron Model
In a metal the outer electrons of each atom are not bound to a single nucleus. They become delocalised and form a “sea of electrons” that can move freely throughout the crystal lattice.
Atoms are arranged in a regular lattice of positively charged ions.
Delocalised electrons are not attached to any particular ion.
The electron sea allows charge to be transferred when an electric field is applied.
Suggested diagram: A lattice of positive ions with a cloud of free electrons moving under an applied electric field.
2. How Conduction Occurs
When a potential difference \$V\$ is applied across a metal, an electric field \$\mathbf{E}\$ is established. The free electrons experience a force \$-e\mathbf{E}\$ and acquire a small average drift velocity \$v_d\$ opposite to the field direction.
The current density \$\mathbf{J}\$ is given by
\$\mathbf{J}=n e \mathbf{v}_d\$
where
\$n\$ = number of free electrons per unit volume (≈ \$10^{28}\,\text{m}^{-3}\$ for typical metals)
Integrating over the cross‑sectional area \$A\$ of the conductor gives the macroscopic current:
\$I = J A = n e A v_d\$
3. Ohm’s Law from the Electron Model
The electric field in a uniform conductor of length \$L\$ is \$E = V/L\$. Substituting \$E\$ into the expression for drift velocity (derived from the balance of electric force and scattering) leads to the linear relationship
\$V = I R\$
where the resistance \$R\$ is
\$R = \rho \frac{L}{A}\$
and the resistivity \$\rho\$ is related to the microscopic properties of the metal:
\$\rho = \frac{m}{n e^2 \tau}\$
with \$m\$ the electron mass and \$\tau\$ the average time between collisions.
4. Factors Influencing Conductivity
Number of free electrons (\$n\$) – More free electrons → lower \$\rho\$, higher conductivity.
Collision time (\$\tau\$) – Fewer collisions (e.g., at lower temperature) increase \$\tau\$, reducing \$\rho\$.
Cross‑sectional area (\$A\$) – Larger \$A\$ reduces \$R\$ for a given length.
Length (\$L\$) – Longer conductors have higher \$R\$.
Material type – Metals have many free electrons; insulators have few or none.
5. Typical Resistivity \cdot alues
Material
Resistivity \$\rho\$ (Ω·m)
Typical Conductivity \$\sigma\$ (S·m⁻¹)
Copper
1.68 × 10⁻⁸
5.96 × 10⁷
Aluminium
2.82 × 10⁻⁸
3.55 × 10⁷
Silver
1.59 × 10⁻⁸
6.30 × 10⁷
Iron
9.71 × 10⁻⁸
1.03 × 10⁷
Glass (insulator)
≈ 10¹⁰ – 10¹⁴
≈ 10⁻¹⁰ – 10⁻¹⁴
6. Comparison with Insulators
In insulators the outer electrons remain tightly bound to their atoms. Consequently:
Very few free charge carriers (\$n\$ ≈ 0).
Large resistivity \$\rho\$ and negligible conductivity \$\sigma\$.
No measurable current flows under ordinary voltages.
7. Summary of Key Points
Metallic conduction is due to a sea of delocalised electrons.
Current arises from the drift of these electrons under an electric field.
Ohm’s law (\$V = IR\$) follows from the linear relationship between drift velocity and electric field.
Resistivity depends on the number of free electrons and how often they scatter.
Increasing cross‑section or decreasing length reduces resistance.
8. Practice Questions
A copper wire 2.0 m long has a cross‑sectional area of \$1.0\times10^{-6}\,\text{m}^2\$. Calculate its resistance. (Use \$\rho_{\text{Cu}} = 1.68\times10^{-8}\,\Omega\!\cdot\!m\$.)
Explain why the resistance of a metal typically decreases when it is cooled.
Compare the conduction mechanisms in a metal and a semiconductor.
Cooling reduces lattice vibrations, increasing the average time between electron collisions (\$\tau\$). From \$\rho = m/(n e^2 \tau)\$ a larger \$\tau\$ gives a smaller resistivity, so \$R\$ decreases.
In a metal, conduction is due to free electrons that are always present. In a semiconductor, electrons must be excited across a band gap; conduction involves both electrons in the conduction band and holes in the valence band, and it is strongly temperature dependent.